Problem: Let $f(x, y) = \ln(x^2) + y$ and $g(t) = (\sin(t), -\cos(t))$. $h(t) = f(g(t))$ $h'(t) = $
Solution: Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(t) = \nabla f(g(t)) \cdot g'(t)$. $\begin{aligned} &g(t) = (\sin(t), -\cos(t)) \\ \\ &g'(t) = (\cos(t), \sin(t)) \\ \\ &\nabla f = \left( 2x \dfrac{1}{x^2}, 1 \right) = \left( \dfrac{2}{x}, 1 \right) \\ \\ &\nabla f(g(t)) = \left( \dfrac{2}{\sin(t)}, 1 \right) \end{aligned}$ Substituting: $\begin{aligned} h'(t) &= \left( \dfrac{2}{\sin(t)}, 1 \right) \cdot (\cos(t), \sin(t)) \\ \\ &= 2 \dfrac{\cos(t)}{\sin(t)} + \sin(t) \end{aligned}$ Answer $h'(t) = 2 \dfrac{\cos(t)}{\sin(t)} + \sin(t)$